Originally Posted by jtt
And the air flowing around the wings gets pushed downwards, producing the lift that keeps you up against gravity. Whenever something is made to move energy is needed (if you accelerate something from zero speed to a speed of v the energy required to do so is half its mass times v squared, i.e. the kinetic energy it obtains). (...) And then there's the so called "induced drag" which results from the wings pushing air downwards.
Hi jtt, excellent explanation!
It just reminds me of an analogy that I found enlightening at the time. A parcel of air with mass m that is pushed downwards at a speed v represents a momentum of m * v, which equals its contribution to the lift
force and, as you say, that requires an
energy of 0,5 * m * v-squared. The
power ( = energy per unit time) required to produce the lift force is "induced drag" force times the airplane velocity. That energy-per-unit-time should equal the energy required for "pushing air downwards". An equation that is not easy to solve, because the air mass involved is infinitely large, the 'induced' velocity asymptotically approaching zero with increasing distance from the airplane. However, if the induced drag corresponds to an elliptical lift distribution over the wing span, and if one crudely assumes that the "air mass" (per unit time) is that passing through a circular area normal to the free stream with a diameter equal to the wing span, and that it is pushed downwards at a uniform 'induced' velocity 'v', then it can be shown that the two energies (per unit time) are equal.
regards,
HN39